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3 years ago

Here is to the people who need points (again) Have a good day!

Engineering

2 answers:

alexdok [17]3 years ago

8 0

Thank you so much!! You too!

g100num [7]3 years ago

7 0

Answer:

SERIOUSLY! TY

Explanation:

I don't think you understand how much this is gonna help me out :) Have an amazing summer! Stay safe!

You might be interested in

What Type of diploma do you need in order To the get into JMU

AysviL [449]

I think a high school diploma should be fine but look more into it on the website https://www.jmu.edu/registrar/students/diploma_main.shtml

3 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Cint%5Climits%5Ea_b%20%7B7x%7D%20%5C%2C%20dx" id="TexFormula1" title="\int\limits^a_b {7x} \

Dafna11 [192]

Answer:

$\frac{7}{2}a^2 - \frac{7}{2}b^2$

Explanation:

$\int\limits^a_b {7x} \, dx$

$=[\frac{7}{2}x^2]^a_b$

$=\frac{7}{2}a^2 - \frac{7}{2}b^2$

5 0

3 years ago

A spherical, stainless steel (k 16 W m1 K-1) tank has a wall thickness of 0.2 cm and an inside diameter of 10 cm. The inside sur

SOVA2 [1]

Answer:

the rate of heat loss is 2.037152 W

Explanation:

Given data

stainless steel K = 16 W $m^{-1}K^{-1}$

diameter (d1) = 10 cm

so radius (r1) = 10 /2 = 5 cm = 5 × $10^{-2}$

radius (r2) = 0.2 + 5 = 5.2 cm = 5.2 × $10^{-2}$

temperature = 25°C

surface heat transfer coefficient = 6 6 W $m^{-2}K^{-1}$

outside air temperature = 15°C

To find out

the rate of heat loss

Solution

we know current is pass in series from temperature = 25°C to 15°C

first pass through through resistance R1 i.e.

R1 = ( r2 - r1 ) / 4 $\pi$ × r1 × r2 × K

R1 = ( 5.2 - 5 ) $10^{-2}$ / 4 $\pi$ × 5 × 5.2 × 16 × $10^{-4}$

R1 = 3.825 × $10^{-3}$

same like we calculate for resistance R2 we know i.e.

R2 = 1 / ( h × area )

here area = 4 $\pi$ r2²

area = 4 $\pi$ (5.2 × $10^{-2}$ )² = 0.033979

so R2 = 1 / ( h × area ) = 1 / ( 6 × 0.033979 )

R2 = 4.90499

now we calculate the heat flex rate by the initial and final temp and R1 and R2

i.e.

heat loss = T1 -T2 / R1 + R2

heat loss = 25 -15 / 3.825 × $10^{-3}$ + 4.90499

heat loss = 2.037152 W

8 0

3 years ago

What gage pressure does a skin diver experience when they dive to 35 ft in the ocean with a water temperature of 55 °F? Report y

TEA [102]

Answer:

p=15.097lbf/in^2

Explanation:

the manometric pressure is that in which the atmospheric pressure is not taken into account, so for this case only the pressure exerted by the water on the bus is calculated using the following equation.

P=ρgh

where

ρ=density of water at 55°F=999.4kg/m^3

g=9.81m/s^2

h=35ft=10.668m

Solving

P=(994.4)(9.81)(10.668)=104067.02Pa = 15.097lbf/in^2

the gage pressure does a skin diver experience when they dive to 35 ft is 15.097lbf/in^2

4 0

3 years ago

Some_____

Thepotemich [5.8K]

Answer:

it’s IGS

Explanation:

because i read

5 0

3 years ago

Here is to the people who need points (*again*) Have a good day!

Here is to the people who need points (again) Have a good day!